Choose independently two numbers \(B\) and \(C\) at random from the interval [0,1] with uniform density. Prove that \(B\) and \(C\) are proper probability distributions. Note that the point \((B, C)\) is then chosen at random in the unit square.
Find the probability that
(a) \(B + C < \frac{1}{2}\)
For \(B + C\) to be less than \(0.5\), \(B\) would have to be less than \(0.5\) and \(C\) would have to be less than \(0.5\). Given that \(B\) is uniform within the interval [0,1], \(P(B<\frac{1}{2}) = 0.5\). Likewise, given that \(C\) is uniform within the interval [0,1], \(P(B<\frac{1}{2}) = 0.5\). Therefore, the probability that both \(B\) and \(C\) are less than \(0.5\) is \((0.5)^2 = 0.25\). Limiting our sample set of \(B\) and \(C\) to situations where both are \(0.5\) or less, we know that \(B + C\) is within the interval [0,1], and we know that half the time \(B + C\) will be over \(0.5\), and half the time it will be under \(0.5\); that is, \(P((B + C)<\frac{1}{2}) = \frac{1}{2}\) given that \(B < 0.5\) and \(C < 0.5\). Since the probability that both \(B\) and \(C\) are less than \(0.5\) is \((0.5)^2 = 0.25\), and \(P((B + C)<\frac{1}{2}) = \frac{1}{2}\) given that \(B < 0.5\) and \(C < 0.5\), we can compute the probability that \(B + C < \frac{1}{2}\) as \(0.25\) * \(0.5\) = \(0.125\).
(b) \(BC < \frac{1}{2}\)
If \(B < \frac{1}{2}\), then \(BC < \frac{1}{2}\). Likewise, if $C < , then \(BC < \frac{1}{2}\). If we think of a square on an \(x/y\) axis that goes from the origin to (0,1), then from (0,1) to (1,1), then from (1,1) to (1,0) and back to the origin as the universe of potential \(B\) and \(C\) values (think of \(B\) as \(x\) and \(C\) as \(y\)), then the fact that we know that if \(B < 0.5\), then \(BC < \frac{1}{2}\), tells us every B value from \(0\) to \(0.5\), or half the square, satisfies the condition, but where \(B > 0.5\), it depends on \(C\). Since we need \(xy < 0.5\) (or \(BC < 0.5\)), we know that the area under the curve \(xy = 0.5\) (or \(y = \frac{1}{2x}\)) where \(0.5 < x < 1\) added to \(0.5\) will give us the probability that \(BC < \frac{1}{2}\). Or:
\(\int_{0.5}^{1} \frac{1}{2x} dx = \frac{ln(|1|)}{2} - \frac{ln(|0.5|)}{2}\)
Computing the integral above and adding 0.5, we get the probability that \(BC < \frac{1}{2}\):
(log(1)/2) - (log(0.5)/2) + 0.5## [1] 0.8465736(C) \(|B-C| < \frac{1}{2}\)
Treating \(B\) as \(x\) and \(C\) as \(y\), then setting \(x - y = 0.5\) or \(x - y = -0.5\) and solving for \(y\), we get \(y = x - 0.5\) and \(y = x + 0.5\). Integrating, we get \(\int_{0}^{-0.5} x - \frac{1}{2} dx = 0.375\) and \(\int_{-0.5}^{0} x + \frac{1}{2} dx = 0.375\). Add the two together and you get \(0.75\).
(d) \(max\{B, C\} < \frac{1}{2}\)
\(P(B<\frac{1}{2}) = 0.5\) and \(P(C<\frac{1}{2}) = 0.5\), so the probability that both are \(<\frac{1}{2}\) is \(0.5^2 = 0.25\). The only way the maximum of \(B\) and \(C\) are not greater than \(0.5\) is if both are less than \(0.5\), which is the compliment of both being less than \(0.5\), or \(1 - 0.25 = 0.75\).
(e) \(min\{B, C\} < \frac{1}{2}\)
For the minimum of \(B\) or \(C\) to be \(<\frac{1}{2}\), both \(B\) and \(C\) must be \(<\frac{1}{2}\). Because the probability that \(P(B<\frac{1}{2}) = 0.5\) and \(P(C<\frac{1}{2}) = 0.5\), the probability that both are \(<\frac{1}{2}\) is \(0.5^2 = 0.25\).